MOLE-MOLE AND MASS-MASS CONVERSIONS

Stoichiometry 1

Using Chemical Equations

Chemical equations are like recipes: Tells chemists what and how much goes in/comes out of a

reaction Given a balanced chemical equation & the quantity of one

substance in the equation, the quantity of any other substance in the same reaction can be calculated

Quantity = the amount of a substance measured in grams, liters, molecules, or moles.

Using Chemical Equations

Balanced Chemical Equations(BCEs) show the ratio of chemical compounds in a reaction

Ratio stays the same when referring to: Single molecules Molar quantities

The above equation can be read as: 1 molecule of nitrogen gas reacts with 3 molecules of hydrogen to

produce 2 molecules of ammonia 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2

moles of ammonia**Remember, 1 mole = 6.02x1023 molecules!**

N2 + 3H2 → 2NH3

Stoichiometry

Stoichiometry – (stoy-key-ometry)The calculation of quantities of substances involved in chemical equations.Balanced chemical equations show molar ratiosAll quantities must be converted to moles FIRST

before using the molar ratio

Mole-Mole Stoichiometry

Consider the following BCE:4Al(s) + 3O2(g) 2Al2O3(s)

If you only had 1.8 moles of Al, how much product could you make?Given: 1.8 mol AlWant: ??? mol Al2O3

Conversion factor: 4 mol Al produce 2 mol Al2O3

1.8 mol Al x2 mol Al2O3

4 mol Al= 0.90 mol Al2O3

Mole-Mole Stoichiometry

If you wanted to produce 24 moles of Al2O3, how many moles of each reactant would you need?

4Al(s) + 3O2(g) 2Al2O3(s)Given: 24 mol Al2O3

Want: ??? mol Al

Want: ??? mol O2

Conversion factors: 4 mol Al produce 2 mol Al2O3, or

3 mol O2 produce 2 mol Al2O3

4 mol Al2 mol Al2O3

3 mol O2

2 mol Al2O3

Mole-Mole Stoichiometry

If you wanted to produce 24 moles of Al2O3, how many moles of each reactant would you need?

4Al(s) + 3O2(g) 2Al2O3(s)

24 mol Al2O3 x

24 mol Al2O3 x

4 mol Al

2 mol Al2O3

3 mol O2

2 mol Al2O3

= 48 mol Al

= 36 mol O2

Mole-Mole Stoichiometry

You can always do mole-mole conversions if you have a BALANCED chemical equation.Unbalanced equations are not useful for stoichiometry.

moles ofsubstanc

e A

moles ofsubstanc

e BBCE

Mole-Mole Stoichiometry Practice

Octane combusts in the presence of oxygen via the following reaction: 2C8H18 + 25O2 16CO2 + 18H2O

How many moles of oxygen gas can react with 7.50 moles of octane?

7.50 mol C8H18 x

If 48 moles of CO2 are produced during one such reaction, how many moles of H2O are also produced?

48 mol CO2 x

25 mol O2

2 mol C8H18

18 mol H2O

16 mol CO2

= 93.8 mol O2

= 54 mol H2O

Mass-Mass Stoichiometry

Moles are not measured directlymass can be measured with a scaleMass (in grams) can be converted to moles and vice-versaMass-mass conversions are more common and practical in real life.

Given the mass of one reactant or product, the mass of any other reactant or product in the reaction can be calculatedRequires a BCEthe unknown can be either a reactant or a product.

Mass-Mass Stoichiometry

Acetylene gas (C2H2) is produced by adding water to calcium carbide (CaC2):CaC2 + 2H2O Ca(OH)2 + C2H2

How many grams of C2H2 are produced by adding water to 5.00 grams of CaC2?Given: 5.00 g CaC2

Want: ??? g C2H2

Conversion factors: 1 mol CaC2 produces 1 mol C2H2

1 mol CaC2 = 64.099 g CaC2 (molar mass of CaC2)

1 mol C2H2 = 26.038 g C2H2 (molar mass of C2H2)

26.038 g C2H2

1 mol C2H2 x

1 mol C2H2

1 mol CaC2 x

1 mol CaC2

64.099 g CaC2 x

Mass-Mass Stoichiometry

Make a plan!Your first step should always be to change to moles!Next, do a mole-mole conversion using the BCE to guide you.Finally, convert from moles back to grams.

5.00 g CaC2 mol CaC2 mol C2H2 g C2H2

5.00 g CaC2 = 2.03 g C2H2

Mass-Mass Stoichiometry

mass A

moles A

moles B

mass B

MMA MMBBCE

Mass-Mass Stoichiometry

Copper metal reacts with concentrated nitric acid to produce copper(II) nitrate, nitrogen dioxide gas, and water:Cu + 4HNO3 Cu(NO3)2 + 2NO2 + 2H2O

How many grams of HNO3 are needed to react completely with 0.450 grams of Cu?Given: 0.450 g CuWant: ??? g HNO3

Conversion factors: 1 mol Cu reacts with 4 mol HNO3

1 mol Cu = 63.546 g Cu (molar mass of Cu) 1 mol HNO3 = 63.012 g HNO3 (molar mass of HNO3)

How many grams of HNO3 are needed to react completely with 0.450 grams of Cu?Cu + 4HNO3 Cu(NO3)2 + 2NO2 + 2H2O

Make a plan!0.450 g Cu mol Cu mol HNO3 g HNO3

0.450 g Cu 1 mol Cu

63.546 g Cu x

4 mol HNO3

1 mol Cux

63.012 g HNO3

1 mol HNO3 x

Mass-Mass Stoichiometry

= 1.78 g HNO3

A Few Reminders

Stoichiometry problems can involve reactants, products, or both.They can also involve moles, grams, liters or milliliters, atoms,

molecules, or any other unit that measures the amount of a particular chemical.

All stoichiometry problems have a mole-mole conversion at their center.At the very least, you’ll need a balanced chemical equation in order

to solve any stoichiometry problem.

It’s a good idea to write not only the unit (g, mol, L, etc) but also the chemical formula when doing conversions.Ex: write 48.0 g Cu(NO3)2, not just 48.0 g.